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                                       Vectors
  AS Exam Questions................................................................................ 1
   Example 9.............................................................................................. 1
   Example 10............................................................................................ 3
   Example 11............................................................................................ 4
   Example 12............................................................................................ 7
   Example 13.......................................................................................... 10

AS Exam Questions
A few basic principles are used over and again in M1 exam
questions. Hopefully the solutions given below should give a few
ideas as to what might appear. Examiner Reports usually state
that vectors is the worst answered question on the paper, but
with a little thought half marks is easily attainable.

Example 9
A particle B moves with constant acceleration (3i + 7j)ms-2. At
time t its velocity is v ms-1. When t = 0, v = (12i – 14j)ms-1.
a)   find the time when B is moving parallel to the vector i.
b)   find the speed of B when t = 8.
c)   find the angle between the direction of motion of B and the
vector i when t = 8.

a)   The particle will be moving parallel to the vector i when it
has no j component.

The velocity at any time is given by:

                         Velocity = initial velocity + ( acceleration × time )

                         = (12i – 14j) + t(3i + 7j)                    (1)
                                                                  Page 2


We are only interested in the j component and particularly when
it is zero:
                -14j + 7tj = 0

Therefore:               t=2

b)    By substituting t = 8 into equation (1) we can find the
velocity of the particle. Speed is the magnitude of the velocity.

                Velocity = (12i – 14j) + t(3i + 7j)

When t = 8               Velocity = (12i – 14j) + 8 × (3i + 7j)

                = (36i + 42j) ms-1

                Speed = 362 + 422 = 55.3ms-1

c)   The velocity at any time gives the direction of motion.



                                       42j

                     θ
                            36i

Therefore the direction is given by:

                             42 
                θ = Tan − 1     
                             36 

                θ = 49.4°
                                                                    Page 3



Example 10
A particle has position vector (3i + 7j) and is moving with speed
39ms-1 in the direction (12i – 5j). Find the position vector at time
t = 5 and the distance from the point (3,4) at this time.

The hint in the question is that the velocity of the particle must
be a multiple of (12i – 5j)ms-1. The magnitude of this vector is 13,
but the speed of the particle is 39ms-1, therefore the velocity of
the particle must be (36i – 15j)ms-1. This one little trick is often
used but if you don’t spot it the question is almost impossible.

The position vector of the particle at any time is given by:

                         Position = initial position + (velocity × time)

                         r = (3i + 7j) + t(36i – 15j)
When t = 5
                         r = (183i – 68j)

The distance between the point and the particle can be found by
Pythagoras theorem.
             Y
                 (3,4)
                                                   X




                   72




                             180                (183,-68)




                   Distance = 722 + 1802 = 193.9m
                                                                   Page 4


The final two examples are of the more challenging type. Setting
the equations for the position vectors is pretty straight forward
as long as you follow the rule:

                 r = initial position + (velocity × time)

In most cases the ships or aeroplanes are on a collision course and
therefore you must set the equations equal to each other and
equate coefficients. These questions may carry in excess of 10
marks but you should be able to make a start.

Example 11
A command post O monitors the movement of two of its ships in
the Gulf. At 1200 hrs a battleship (B) has position (-2i + 10j) km
relative to O and has constant velocity of (3i + 2j) kmh-1. A
frigate (F) is at the point with position vector (4i + 5j) km and has
constant velocity (-3i + 7j) kmh-1, where i and j are unit vectors
directed due east and due north respectively.
a)    The captain of one ship has been taken ill, show that the
two ships will collide.
The command post contacts the battleship and orders it to
reduce its speed to move with velocity (2i + 2j) kmh-1.
                                             r
                                           uuu
b)    Find an expression for the vector BF at time t hours after
noon.
c)    Find the distance between B and F at 1400 hrs.
d)    Find the time at which F will be due north of B.

a)   The position of each ship is given by it’s position vector:

                 position = initial position + (velocity × time)
                                                                 Page 5


So for the battleship:

                 rb = (-2i + 10j) + t(3i + 2j)

And for the frigate:

                 rf = (4i + 5j) + t(-3i + 7j)

If the two ships are to collide then for some value of t their
respective i and j components must be equal.

Therefore by equating i’s:

                 -2 + 3t = 4 -3t

                 t=1

Substituting the value of t = 1 into rb and rb gives the same
position vector of (i + 12j). Therefore the two ships will collide
after one hour at the point with position vector (i + 12j).

b)   The position vector for the battleship must change to take
account of the new velocity:

                 rb = (-2i + 10j) + t(2i + 2j)
                                              r
                                            uuu
We have been asked to find the vector BF as shown in the
diagram below.
                                                                      Page 6


              N

                             B



                                        F



              O                                E


By triangle law:
                   uur uur uuur
                   0B + BF = OF

                         r
                   uur uuu uur
                   BF = OF − 0B

        uur              r
                       uuu
Where 0B = rb and OF = rf
                           r
                         uuu
Therefore:               BF = rf – rb


                   = (4i + 5j) + t(-3i + 7j) – ((-2i + 10j) + t(2i + 2j))
                     r
                   uuu
                   BF = (6i - 5j) + t(-5i + 5j)

                               r
                             uuu
c)    The magnitude of BF gives the distance between the two
ships, at 1400 hrs, t = 2

                   = (6i - 5j) + 2(-5i + 5j)

                   = (-4i + 5j)

                   Magnitude =     41Km
                                                                Page 7

                                      r
                                    uuu
d)   If F is due north of B, then BF will have no i component.
                   r
                 uuu
                 BF = (6i - 5j) + t(-5i + 5j)


                 6i -5ti = 0

                 t = 1hr 12mins

Example 12
Two cars P and Q are moving on straight horizontal roads with
constant velocities. The velocity of P is 25ms-1 due east and the
velocity of Q is (10i + 8j)ms-1. Initially P is at rest at the origin
and Q has the position vector 230im relative to the origin. At
time t seconds the position vectors of P and Q are r metres and s
metres respectively.
a)    Find expressions for r and s in terms of t.
                                   r
                                 uuu
b)    Write an expression for PQ .
c)    Find the time when the bearing of Q from P is 045º
d)    Find the time when the cars are 200m apart.

a)   The position vector for car P is given by:

                 r = initial position + (velocity × time)

                 r = 0 + 25ti

Therefore for Q:

                 s = 230i + t(10i + 8j)
                                                                   Page 8


b)   Using the triangle law:
                 y

                                      P



                                                 Q



                 O                                       x
                  r    r
                uu uuu uuuu r
                0P + PQ = OQ

                  r    r
                uuu uuuu uur
                PQ = OQ − 0P

       uur           r
                  uuuu
Where 0P = r and OQ = s
                           r
                         uuu
Therefore:               PQ = s – r


                = 230i + t(10i + 8j) – 25ti

                = 230i + t(8j – 15i)
                                                               r
                                                             uuu
c)   If the bearing of Q from P is 045º, then the vector PQ
must be parallel to the vector (i + j).

Therefore
                230i + t(8j – 15i) = m(i + j)

Equating coefficients:

                230 - 15t = m         (1)       8t = m         (2)
                                                           Page 9


Substituting gives:

                230 - 15t = 8t

                t = 10

The bearing of Q from P is 045º after 10 seconds.
                              r
                            uuu
d)   The magnitude of PQ will give the distance that the two
                                               r
                                             uuu
cars are apart. We need the value of t where PQ = 200
            r
          uuu
Rewriting PQ gives:
                  r
                uuu
                PQ = (230 – 15t)i + 8tj

                   r
                 uuu
                 PQ =      (230 − 15t)2 + 64t2

                   r
                 uuu
                 PQ =      52900 − 6900t − 225t2 + 64t2


                2002 = 52900 − 6900t − 161t2


                161t2 + 6900t − 12900 = 0


                t = 1.79
                                                                     Page 10



Example 13
At 1200 hrs the position vectors of two helicopters A and B are
rA and rB as outlined below. Use the velocity vectors vA and vB to
give the position vectors of A and B at a time t hours after noon.

rA = (3i + 5j)    rB = (5i + 2j)            vA = (i + 2j)      vB = (2i +
3j)

Find an expression at time t hours after noon for the position
vector of B relative to A.

If d is the distance in Km between the two helicopters find the
value of d2 in terms of t.

Find the time at which the helicopters are closest together. Give
the value of the minimum distance.

a) For Helicopters A and B the position vectors at time t hours
after noon are:

      rA = (3i + 5j) + t(i + 2j)            rB = (5i + 2j) + t(2i + 3j)

b)    The position vector of B relative to A is given by:

                  rB – rA = 2i – 3j + t(i + j)

                  = (2 + t)i + (t - 3)j
                                                              Page 11


c)   The magnitude of the vector in part (b) gives the distance
between the two particles.
Therefore:

                d2 = (2 + t)2 + (t – 3) 2

                d2 = 2t2 - 2t + 13

By completing the square we can find the minimum value:

                d2 = 2[t2 - t + 6.5]

Remembering to halve the t coefficient:
                                       (-0.5)2 = 0.25 we need 6.5
               d2 = 2[(t – 0.5)2 + 6.25]

The minimum value occurs when t = 0.5. Therefore the minimum
distance between the two helicopters is √12.5 Km

				
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